Phosphorus pentachloride decomposes according to the chemical equation

PCl5(g) -----> PCl3 (g)+ Cl2(g)
Kc= 1.80 at 250 degrees C

A 0.463 mol sample of PCl5(g) is injected into an empty 4.80 L reaction vessel held at 250 °C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.

PLEASE HELP!!!! Can you confirm the correct answer as well. I have one more attempt. Thanks

M PCl5 = 0.463 mols/4.80L = 0.09646 which is more places than allowed but I like to keep one extra place, go through the calculations and round in the last step.

..........PCl5 ==>PCl3 + Cl2
initial.0.09646......0......0
change...-x.........x.......x
equil..0.09646-x.....x......x

Kc = 1.80 = (PCl3)(Cl2)/(PCl5)
Substitute from the ICE chart into the Kc expression and solve for x, then for 0.09646

What answers did u get for PCl5 and PCl3...I want to verify them please.

I didn't work it. Give me your answers and I will check it.

I got x=0.091766....now what do i do? I'm confused. Also can check if this is right for x?

I obtained 0.0918M for x and that is (PCl3). The problem also asked for (PCl5) and that is 0.09646-x = ? and round to 3 s.f. I obtained 0.00466 for that.

yes it is correct! thank you :) I really appreciate it!

How do you solve for x with the equation 1.8 = x^2 / .0964 - x? i don't understand what to do isolate x

I want to know that too ^

That always stumps me on the hw!

you cannot isolate X by its self without using the quadratic equation:

x = -B +or- square-root(B^2 - 4AC)
------------------------------
2A

To calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium, we can use the equilibrium constant (Kc) and the initial moles of PCl5 injected into the reaction vessel.

First, let's calculate the initial concentration of PCl5(g):
Initial moles of PCl5(g) = 0.463 mol
Initial volume of the reaction vessel = 4.80 L
Initial concentration of PCl5(g) = moles/volume = 0.463 mol / 4.80 L

Next, let's determine the concentrations at equilibrium using the equilibrium constant (Kc):
Kc = [PCl3(g)][Cl2(g)] / [PCl5(g)]

Let's assume that at equilibrium, the moles of PCl5(g) that reacted is 'x'. Thus, the moles of PCl3(g) and Cl2(g) formed will also be 'x'.

At equilibrium, the moles of PCl5(g) will be (0.463 - x) and the volume of the reaction vessel will remain constant, i.e., 4.80 L.

Therefore, the concentration of PCl5(g) at equilibrium = (0.463 - x) mol / 4.80 L
The concentration of PCl3(g) at equilibrium = x mol / 4.80 L

Now, we can substitute these concentration values into the equilibrium expression:

Kc = [PCl3(g)][Cl2(g)] / [PCl5(g)]
1.80 = (x / 4.80) * (x / 4.80) / [(0.463 - x) / 4.80]

Simplifying the equation:

1.80 = (x^2) / [(0.463 - x)]
1.80 * (0.463 - x) = x^2

Rearranging:

1.80 * 0.463 - 1.80x = x^2
0.834 - 1.80x = x^2
x^2 + 1.80x - 0.834 = 0

Now, we can solve this quadratic equation to find the value of 'x'. We can use the quadratic formula:

x = [-b ± sqrt(b^2 - 4ac)] / (2a)

where a = 1, b = 1.80, c = -0.834

Calculating the value of 'x', we find that x = 0.271.

Plugging this value back into the equation for equilibrium concentrations:

Concentration of PCl5(g) at equilibrium = (0.463 - 0.271) mol / 4.80 L
Concentration of PCl3(g) at equilibrium = 0.271 mol / 4.80 L

Simplifying,

Concentration of PCl5(g) at equilibrium = 0.192 mol / 4.80 L
Concentration of PCl3(g) at equilibrium = 0.271 mol / 4.80 L

So, the concentrations of PCl5(g) and PCl3(g) at equilibrium are 0.04 M and 0.06 M, respectively.

Please note that the provided answer is based on the information and calculations provided. Double-check your calculations to ensure accuracy.